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Hollow cylinder: heat transfer; temperature at 10 cm

Short answer

Taking 10 cm as diameter (radius 5 cm), heat transfer per unit length Q/L = 2π×60×(200−50)/ln(0.06/0.04) ≈ 139.5 kW/m. Temperature at r = 5 cm: T = 200 − 150×ln(1.25)/ln(1.5) ≈ 117.4 °C.

A hollow cylinder with inner and outer diameters of 8 cm and 12 cm respectively has an inner surface temperature of 200°C and an outer surface temperature of 50°C. If the thermal conductivity of the cylinder material is 60 W/mK, determine the heat transfer from the unit length of the pipe. Also determine the temperature at the surface at a radial distance of 10 cm from the axis of the cylinder.

Reading of the data. The inner and outer radii are 4 cm and 6 cm, so a radial distance of 10 cm lies outside the wall and cannot be a point in the material. The 10 cm is taken as a diameter, i.e. radius r = 5 cm (the same problem is asked with "5 cm"), which lies inside the wall.

Given: r₁ = 0.04 m, r₂ = 0.06 m, T₁ = 200 °C, T₂ = 50 °C, k = 60 W/m·K. Steady radial conduction, no heat generation.

Heat transfer per unit length

Q/L = 2πk (T₁ − T₂)/ln(r₂/r₁) = 2π × 60 × 150/ln(1.5) = 56 549/0.4055

Q/L ≈ 139466 W/m (≈ 139.5 kW/m)

Temperature at r = 5 cm

T(r) = T₁ − (T₁ − T₂) ln(r/r₁)/ln(r₂/r₁)

T(0.05) = 200 − 150 × ln(1.25)/ln(1.5) = 200 − 150 × 0.22314/0.40547

T ≈ 117.4 °C

(For r ≥ 6 cm the material ends; at r = 6 cm the temperature is 50 °C.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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